<span>a)
Capacitance = k x ε° x area / separation
ε° = 8.854 10^-12 F/ m
k = 2.4max
average k = 0.78 / 1.27 * 2.4 +(1.27- 0.78) / 1.27 * 1 = 1.474 + 0.386 = 1.86
(61.4 % separation k = 2.4 --- 38.6 % k = 1 air --- average k = 0.614 * 2.34 + 0.386 * 1 = 1.86
area = 145 cm2 = 0.0145 m2
separation = 1.27 cm 0.0127 m
C = 1.86 * 8.854 10^-12 * 0.0145 / 0.0127 = 18.8 pF
b) Q = C * V --- 18.8 * 83 = 1560.4 pC = 1.5604 nC
c) E = V / d = 83 / 0.0127 = 6535.4 V/m </span>
Answer:
No
Explanation:
Social media fame isnt worth it, because youre not really able to connect with people. Real life relationships with friends, family, and everything else is way more important. having people notice you or be attracted to you through the internet is nothing compared to real life social interactions. Dont leave people just to be online, youre going to regret it.
Answer:
5.4 ms⁻¹
Explanation:
Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.
= length of the meter stick = 1 m
= mass of the meter stick
= angular speed of the meter stick as it hits the floor
= speed of the other end of the stick
we know that, linear speed and angular speed are related as

= height of center of mass of meter stick above the floor = 
= Moment of inertia of the stick about one end
For a stick, momentof inertia about one end has the formula as

Using conservation of energy
Rotational kinetic energy of the stick = gravitational potential energy

Answer:
Hari didn't plan to go abroad
Explanation:
Abroad planned to go to hari.
Answer:
h = 3.1 cm
Explanation:
Given that,
The volume of a oil drop, V = 10 m
Radius, r = 10 m
We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

So, the thickness of the film is equal to 3.1 cm.